Question

rút gọn (x-y) (x cộng y) ( x^2 cộng y^2) (x^4 cộng y^4 ) (x^8 cộng Y^8 )

Answer 1

Ta có 

\(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

Answer 2

Ta có \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(=\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(=\left(x^8-y^8\right)\left(x^8+y^8\right)\)

\(=x^{16}-y^{16}\)

Answer 3

Đặt \(A=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

        Áp dụng công thức \(\left(a-b\right)\left(a+b\right)=a^2-b^2\) ta được:

\(\Leftrightarrow A=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(\Leftrightarrow A=\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

\(\Leftrightarrow A=\left(x^8-y^8\right)\left(x^8+y^8\right)\)

\(\Leftrightarrow A=\left(x^{16}-y^{16}\right)\)

Answer 4

Ta có \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

=\(\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

=\(\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)=\left(x^8-y^8\right)\left(x^8+y^8\right)\)

=\(x^{16}-y^{16}\)

Answer 5

tớ cũng gioongws các bn trên

Answer 6

mk mới lớp 5 thôi